AQA FP2 de Moivre's theorem question

I know that:

$z^n + 1/z^n = 2cosx$

$z^n - 1/z^n = 2isinx$

For part (b) I get:

$-32cos^3(x)sin^3(x) = Asin(6x) + Bsin(2x)$

Which seems like it'll get the answer to part ii but I don't know how to answer part i from where I am now, or if I'm using the right method to find A and B

For (b) (i) consider the following:

$\begin{aligned} \left( z + \frac{1}{z} \right)^3 \left( z - \frac{1}{z} \right)^3 & = \left[ \left( z + \frac{1}{z} \right) \left( z - \frac{1}{z} \right) \right]^3 \\ & = \left( z^2 - \frac{1}{z^2} \right)^3 \\ & = \displaystyle \sum_{r=0}^3 \binom{3}{r} (z^2)^{3-r} \left(\frac{-1}{z^2} \right)^2 \end{aligned}$

Thanks, but I've never had to use sigma notation in these questions before, is there a way to do it using only trig as I think that's how I'm supposed to answer it?

http://filestore.aqa.org.uk/subjects/AQA-MFP2-TEXTBOOK.PDF (Page 83)

Thanks, but I've never had to use sigma notation in these questions before, is there a way to do it using only trig as I think that's how I'm supposed to answer it?

http://filestore.aqa.org.uk/subjects/AQA-MFP2-TEXTBOOK.PDF (Page 83)

You don't need the sigma notation at all for this since you don't need a full expansion - just use the 1st 2 lines of Khalil's solution then apply part (a).

May I ask what the values of A and B are?

In that case the textbook is wrong...

The textbook isn't wrong. They too have -3 as the value of B:



In any case, I strongly believe that you're not required to use any complex numbers for (b) (i). This is because (b) (ii) tells you to substitute $z = \cos + i\sin \theta$ in the identity for (b) (i) with the values of A and B that have already been found.

Have you tried using the binomial expansion I recommended in my first post?